Question

Como simplificar as expressões:
a - (3x-1)²+(x-2)²
b- (a³-2)²-(a³+1)²
c- (1-2x)²- (1+x)²

Answer 1

a)
(3x-1)² + (x-2)²=

9x²-6x+1+x²-4x+4=
10x²-10x+5

b)
(a³-2)²-(a³+1)²=

$a^{6} -4 a^{3} +4-( a^{6} +2 a^{3} +1)=$

$a^{6} -4 a^{3} +4- a^{6} -2 a^{3} -1=$

-6a³+3

c)
(1-2x)²-(1+x)²=

1-4x+4x²-(1+2x+x²)=
1-4x+4x²-1-2x-x²=

3x²-6x

Answer 2

a) $(3x-1)^2+(x-2)^2$

Em geral, $(a+b)=a^2+2ab+b^2$ e $(a-b)^2=a^2-2ab+b^2$.

Assim:

$\rhd$ $(3x-1)^2=(3x)^2-2\cdot 3x\cdot 1+1^2=9x^2-6x+1$

$\rhd$ $(x-2)^2=x^2-2\cdot x\cdot 2+2^2=x^2-4x+4$

Portanto, :

$(3x-1)^2+(x-2)^2=(9x^2-6x+1)+(x^2-4x+4)=10x^2-10x+5$.

b) De modo análogo, obtemos:

$\bullet$ $(a^3-2)^2=(a^3)^2-2\cdot a^3\cdot 2+2^2=a^6-4a^3+4$

$\bullet$ $(a^3+1)^2=(a^3)^2+2\cdot a^3\cdot1+1^2=a^6+2a^3+1$

Logo, :

$(a^3-2)^2-(a^3+1)^2=(a^6-4a^3+4)-(a^6+2a^3+1)=-6a^3+3$

c) Por fim, temos:

$\rhd$ $(1-2x)^2=1^2-2\cdot 1\cdot 2x+(2x)^2=1-4x+4x^2$

$\rhd$ $(1+x)^2=1^2+2\cdot1\cdot x+x^2=1+2x+x^2$

E obtemos, :

$(1-2x)^2-(1+x)^2=(1-4x+4x^2)-(1+2x+x^2)=3x^2-6x$